Internal transformer composite-defect fuzzy diagnostic method based on gas dissolved in oil

ABSTRACT

A transformer internal composite defect fuzzy diagnosis method based on gas dissolved in oil, comprising: a step of acquiring monitoring data of volume concentrations of five types of monitored feature gas; a step of determining ratio codes; a step of modifying a three-ratio method; a step of fuzzifying a boundary range; a step of calculating probabilities of the ratio codes; a step of calculating a probability of occurrence of each defect fault; and finally obtaining a fault type of a transformer. The method has the beneficial effects that: the method is simple and easy to achieve, and particularly suitable for being applied to an on-line transformer state monitoring system; based on a concept of fuzzy logic, diagnosis of composite defects of the transformer under a complicated state and evaluation of the degree of severity can be achieved, and the problem of sudden change caused by criterion boundary absolutisation can be effectively avoided; and multi-feature information such as an attention value and a ratio of the gas dissolved in the oil are merged and analysed, thereby effectively improving the diagnosis reliability.

FIELD OF THE INVENTION

The present invention relates to the technical field of powertransformer fault diagnosis, in particular to a fuzzy diagnosis methodfor transformer internal composite defect based on gas dissolved in oil.

BACKGROUND OF THE INVENTION

Power transformers are important devices in electric power systems andare of great significance for safe and reliable operation of theelectric grid. Routine test of power transformers is an important meansfor detecting potential hazards in the transformers and avoid outburstof accidents. Among the tests, transformer oil chromatogram test is avery effective test and includes rich information on insulation state ofthe device, and can be used to find defects in the transformer such asinternal discharge, local overheat, and moistened insulation, etc.;hence, transformer oil chromatogram test is widely applied in electricpower systems. Accurate diagnosis of internal defects in transformers ishelpful for judging the positions and types of the defects in thedevices, and is always a key subject in the research in the art. Basedon that, a scientific overhaul strategy can be established, and therebythe efficiency of equipment operation, maintenance and overhaul can begreatly improved, and the power distribution reliability of the electricgrid can be improved. At present, classical transformer oil chromatogramanalysis methods mainly include characteristic gas method, Rogers ratiomethod, IEC three-ratio method, Dewey triangular chart method, andmodified three-ratio method, etc. Among those methods, thecharacteristic gas method is a method for rating the defect according tothe concentration of characteristic gases and total hydrocarbonconcentration, and is only suitable for qualitatively judging whetherthere is any defect or not. The Rogers four-ratio method is developed onthe basis of Doerenburg five-ratio method, and utilizes gas ratios ascriteria for judging the types of transformer defects, like the IECthree-ratio method. Such kind of ratio methods only make sense for theratios of defective transformer, but may result in misjudgement undernormal conditions; in addition, these methods may have problems inactual application, for example, there is no corresponding ratio code,the boundaries of criteria are absolutized, and composite defects cannotbe diagnosed accurately, etc. The modified three-ratio method proposedin the Chinese Standard GB/T7252-2001 is developed on the basis of theIEC three-ratio method by correcting corresponding codes according tothe statistical analysis result of transformer data in China, but itstill utilize gas ratios as criteria for judging the types oftransformer defects. The Dewey triangular chart method is a method thatdifferentiate the types of defects on the basis of the triangular chartcoordinates of gas ratio distribution, in which each type of defectscorresponds to a certain region; though this method solves the problemthat there is no corresponding code in the ratio method, but still haveother problems, for example, the boundaries are absolutized andcomposite defects cannot be diagnosed accurately.

In summary, the diagonisis characteristic criteria used in different oilchromatogram diagnosis methods for transformer defects only rely onsimplex characteristic information, such as characteristic gas type, gasconcentration or gas ratio, the boundaries of diagnostic criteria aretoo absolutized, and the diagnosis result cannot reveal the severitiesor probabilities of occurrence of the defects. In actual application,the transformer defects are complex, and are usually composite defects;consequently, such defects cannot be identified with existing diagnosismethods. Therefore, it is necessary to improve the existing oilchromatogram diagnosis methods for transformer internal defects.

CONTENTS OF THE INVENTION

The technical problem to be solved by the present invention is toprovide a fuzzy diagnosis method for transformer internal compositedefect based on gas dissolved in oil, which can effectively solve theproblem that the criteria boundaries are absolutized and compositedefects cannot be diagnosed with conventional analysis method based ongas dissolved in oil, can comprehensively utilize the information of avariety of characteristic quantities, and can effectively improve thereliability of defect and fault diagnosis.

To solve the above-mentioned technical problem, the present inventionemploys the following technical scheme: a fuzzy diagnosis method fortransformer internal composite defect based on gas dissolved in oil,comprising the following steps:

(I) Acquiring monitoring data of volumetric concentration of five typesof monitored characteristic gases, i.e., hydrogen, methane, ethane,ethylene and acetylene; calculating the sum of the volumetricconcentrations of methane, ethane, ethylene and acetylene (i.e.,volumetric concentration of total hydrocarbons) from the monitoringdata; judging whether the monitoring data of the five types ofcharacteristic gases or the volumetric concentration of totalhydrocarbons exceeds an alert value, which is selected as per theChinese Standard GB/T7252-2001; if the monitoring data or the volumetricconcentration of total hydrocarbons exceeds the alert value, furtherdiagnosis is required; in that case, going to step (II); otherwisejudging that the transformer has no defect or fault, if the monitoringdata and the volumetric concentration of total hydrocarbons are normal;

(II) Determining ratio codes:

First, the ratios are set as follows:

${r_{1} = \frac{c_{1}\left( {C_{2}H_{2}} \right)}{c_{2}\left( {C_{2}H_{4}} \right)}},{r_{2} = \frac{c_{3}\left( {CH}_{4} \right)}{c_{4}\left( H_{2} \right)}},{r_{3} = \frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{5}\left( {C_{2}H_{6}} \right)}},{r_{4} = {{\frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{5}\left( {C_{2}H_{6}} \right)} \times \frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{1}\left( {C_{2}H_{2}} \right)}} = {r_{3} \times \frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{1}\left( {C_{2}H_{2}} \right)}}}}$

wherein, c₁(C₂H₂), c₂(C₂H₄), c₃(CH₄), c₄(H₂) and c₅(C₂H₆) respectivelyrepresent the volumetric concentration of five types of characteristicgases (acetylene, ethylene, methane, hydrogen and ethane), in unit ofμL/L;

Then, the ratio codes are determined according to the following rules:

If r₁<0.1, the ratio code of r₁ is 0; if 0.1≦r₁<1, the ratio code of r₁is 1; if 5.1r₁<3, the ratio code of r₁ is 1; if r₁ the ratio code of r₁is 2;

If r₂<0.1, the ratio code of r₂ is 1; if 0.1≦r₂<1, the ratio code of r₂is 0; if 1≦r₂<3, the ratio code of r₂ is 2; if the ratio code of r₂ is2;

If r₃<0.1, the ratio code of r₃ is 0; if 0.1r₃<1, the ratio code of r₃is 0; if 1.5r₃<3, the ratio code of r₃ is 1; if r₃≧3, the ratio code ofr₃ is 2;

If r₄≦1.5, the ratio code of r₄ is 0; if r₄>1.5, the ratio code of r₄ is1;

(III) Correcting the method for determining the types of transformerdefects or faults on the basis of three ratios as specified in theChinese Standard GB/T7252-2001:

Based on the types of transformer defects or faults corresponding to thethree ratio codes specified in the Chinese Standard GB/T7252-2001, aratio code 011 corresponding to the defect or fault type of partialdischarge is added;

A fourth ratio r₄ is added on the basis of the three ratio codes; forthe type of defect or fault with ratio code 101 diagnosed with thethree-ratio method, if r₄≦1.5, the transformer is judged as having aspark discharge defect or fault; if r₄>1.5, the transformer is judged ashaving an arc discharge defect or fault;

Thus, a method for judging the types of transformer defects or faultsaccording to ratio codes is obtained as follows:

If the ratio code of r₁ is 0, the ratio code of r₂ is 1, the ratio codeof r3 is 0, 1 or 2, and the ratio code of r₄ is 0 or 1, the type oftransformer defect or fault is partial discharge;

If the ratio code of r₁ is 0, the ratio code of r₂ is 0, the ratio codeof r₃ is 1, and the ratio code of r₄ is 0 or 1, the type of transformerdefect or fault is low-temperature overheat lower than 300□;

If the ratio code of r₁ is 0, the ratio code of r₂ is 2, the ratio codeof r₃ is 0, and the ratio code of r₄ is 0 or 1, the type of transformerdefect or fault is low-temperature overheat lower than 300□;

If the ratio code of r₁ is 0, the ratio code of r₂ is 2, the ratio codeof r₃ is 1, and the ratio code of r₄ is 0 or 1, the type of transformerdefect or fault is 300-700□ moderate-temperature overheat;

If the ratio code of r₁ is 0, the ratio code of r₂ is 0 or 2, the ratiocode of r₃ is 2, and the ratio code of r₄ is 0 or 1, the type oftransformer defect or fault is high-temperature overheat higher than700□;

If the ratio code of r₁ is 2, the ratio code of r₂ is 0, 1 or 2, theratio code of r₃ is 0, 1 or 2, and the ratio code of r₄ is 0 or 1, thetype of transformer defect or fault is spark discharge;

If the ratio code of r₁ is 1, the ratio code of r₂ is 0, the ratio codeof r₃ is 1, and the ratio code of r₄ is 0, the type of transformerdefect or fault is spark discharge;

If the ratio code of r₁ is 1, the ratio code of r₂ is 0, the ratio codeof r₃ is 1, and the ratio code of r₄ is 1, the type of transformerdefect or fault is arc discharge;

If the ratio code of r₁ is 1, the ratio code of r₂ is 0, 1 or 2, theratio code of r₃ is 0 or 2, and the ratio code of r₄ is 0 or 1, the typeof transformer defect or fault is arc discharge;

If the ratio code of r₁ is 1, the ratio code of r₂ is 1 or 2, the ratiocode of r₃ is 1, and the ratio code of r₄ is 0 or 1, the type oftransformer defect or fault is arc discharge;

(IV) Blurring the boundary ranges of the ratios r₁, r₂, r₃ and r₄ with asemi-Cauchy rising/falling function, and representing the rising edgesand falling edges of the boundaries with the semi-Cauchy rising/fallingfunction as follows:

${\mu_{d}(r)} = \left\{ {{\begin{matrix}1 & {,{r \leq A}} \\\frac{1}{1 + \left( \frac{A - r}{a} \right)^{2}} & {,{others}}\end{matrix}{\mu_{a}(r)}} = \left\{ \begin{matrix}1 & {,{r \geq A}} \\\frac{1}{1 + \left( \frac{A - r}{a} \right)^{2}} & {,{others}}\end{matrix} \right.} \right.$

wherein, μ_(d)(r) is a falling edge function; μ_(a)(r) is a rising edgefunction; A is a boundary parameter; a is a distribution parameter; thevalues of A and a are as follows:

The rising edge boundary parameter of r₁ is 0.08, and the correspondingdistribution parameter is 0.01

The falling edge boundary parameter of r₁ is 3.1, and the correspondingdistribution parameter is 0.1;

The rising edge boundary parameter of r₂ is 0.06, and the correspondingdistribution parameter is 0.02;

The falling edge boundary parameter of r₂ is 0.6, and the correspondingdistribution parameter is 0.2;

The rising edge boundary parameter of r₃ is 0.8, and the correspondingdistribution parameter is 0.1;

The falling edge boundary parameter of r₃ is 3.6, and the correspondingdistribution parameter is 0.3;

The boundary parameter of r₄ is 1.43, and the corresponding distributionparameter is 0.1;

(V) Obtaining the probabilities of the cases that the ratio codes of theratios r₁, r₂ and r₃ are 0, 1 and 2 respectively and the probabilitiesof the cases that the ratio code of r₄ is 0 or 1 respectively, with thesemi-Cauchy rising/falling function; the expressions are as follows:

Probability f-code0(r₁) of the case that the ratio code of r₁ is 0:

${f\text{-}{code}\mspace{11mu} 0\left( r_{1} \right)} = \left\{ \begin{matrix}1 & \left( {r_{1} \leq 0.08} \right) \\\frac{1}{1 + \left( \frac{0.08 - r_{1}}{0.01} \right)^{2}} & \left( {r_{1} > 0.08} \right)\end{matrix} \right.$

Probability f-code1(r₁) of the case that the ratio code of r₁ is 1:

${f\text{-}{code}\; 1\left( r_{1} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{0.08 - r_{1}}{0.01} \right)^{2}} & \left( {r_{1} < 0.08} \right) \\1 & \left( {0.08 \leq r_{1} \leq 3.1} \right) \\\frac{1}{1 + \left( \frac{3.1 - r_{1}}{0.1} \right)^{2}} & \left( {r_{1} > 3.1} \right)\end{matrix} \right.$

Probability f-code2(r₁) of the case that the ratio code of r₁ is 2:

${f\text{-}{code}\; 2\left( r_{1} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{3.1 - r_{1}}{0.1} \right)^{2}} & \left( {r_{1} < 3.1} \right) \\1 & \left( {r_{1} \geq 3.1} \right)\end{matrix} \right.$

Probability f-code0(r₂) of the case that the ratio code of r₂ is 0:

${f\text{-}{code}\; 0\left( {r\; 2} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{0.06 - r_{2}}{0.02} \right)^{2}} & \left( {r_{2} < 0.06} \right) \\1 & \left( {0.06 \leq r_{2} \leq 0.6} \right) \\\frac{1}{1 + \left( \frac{0.6 - r_{2}}{0.2} \right)^{2}} & \left( {r_{2} > 0.6} \right)\end{matrix} \right.$

Probability f-code1(r₂) of the case that the ratio code of r₂ is 1:

${f\text{-}{code}\; 1\left( {r2} \right)} = \left\{ \begin{matrix}1 & \left( {r_{2} \leq 0.06} \right) \\\frac{1}{1 + \left( \frac{0.06 - r_{2}}{0.02} \right)^{2}} & \left( {r_{2} > 0.06} \right)\end{matrix} \right.$

Probability f-code2(r₂) of the case that the ratio code of r₂ is 2:

${f\text{-}{code}\; 2\left( r_{2} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{0.6 - r_{2}}{0.2} \right)^{2}} & \left( {r_{2} < 0.6} \right) \\1 & \left( {r_{2} \geq 0.6} \right)\end{matrix} \right.$

Probability f-code0(r₃) of the case that the ratio code of r₃ is 0:

${f\text{-}{code}\; 0\left( r_{3} \right)} = \left\{ \begin{matrix}1 & \left( {r_{3} \leq 0.8} \right) \\\frac{1}{1 + \left( \frac{0.8 - r_{3}}{0.1} \right)^{2}} & \left( {r_{3} > 0.8} \right)\end{matrix} \right.$

Probability f-code1(r₃) of the case that the ratio code of r₃ is 1:

${f\text{-}{code}\; 1\left( r_{3} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{0.6 - r_{3}}{0.1} \right)^{2}} & \left( {r_{3} < 0.8} \right) \\1 & \left( {0.8 \leq r_{3} \leq 3.6} \right) \\\frac{1}{1 + \left( \frac{3.6 - r_{3}}{0.3} \right)^{2}} & \left( {r_{3} > 3.6} \right)\end{matrix} \right.$

Probability f-code2(r₃) of the case that the ratio code of r₃ is 2:

${f\text{-}{code}\; 2\left( r_{3} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{3.6 - r_{3}}{0.3} \right)^{2}} & \left( {r_{3} > 3.6} \right) \\1 & \left( {0.8 \leq r_{3} \leq 3.6} \right)\end{matrix} \right.$

Probability f-code0(r₄) of the case that the ratio code of r₄ is 0:

${f\text{-}{code}\; 0\left( r_{4} \right)} = \left\{ \begin{matrix}1 & \left( {r_{4} \leq 1.43} \right) \\\frac{1}{1 + \left( \frac{1.43 - r_{4}}{0.1} \right)^{2}} & \left( {r_{4} > 1.43} \right)\end{matrix} \right.$

Probability f-code1(r₄) of the case that the ratio code of r₄ is 1:

${f\text{-}{code}\; 1\left( r_{4} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{1.43 - r_{4}}{0.1} \right)^{2}} & \left( {r_{4} < 1.43} \right) \\1 & \left( {r_{4} \geq 1.43} \right)\end{matrix} \right.$

(VI) Representing the probabilities of ratio codes with maximum valuelogic and minimum value logic, and thereby obtaining a fuzzymulti-valued form of the diagnostic result of the types of transformerdefects or faults; the probabilities of the types of transformer defectsor faults are as follows:

f(partial discharge)=min[f-code0(r ₁), f-code1(r ₂)];

f(low-temperature overheat)=max{min[f-code0(r ₁), f-code0(r ₂),f-code1(r ₃)], min[f-code0(r ₁), f-code2(r ₂), f-code0(r3)]};

f(moderate-temperature overheat)=min[f-code0(r ₁), f-code2(r ₂),f-code1(r3)];

f(high-temperature overheat)=max{min[f-code0(r ₁), f-code0(r ₂),f-code2(r ₃)], min[f-code0(r ₁), f-code2(r2), f-code2(r ₃)]};

f(spark discharge)=max{f-code2(r ₁), min[f-code 1(r ₁), f-code0(r ₂),f-code1(r ₃), f-code0(r ₄)]};

f(arc discharge)=max{min[f-code 1(r ₁), f-code0(r ₂), f-code1(r ₃),f-code1(r ₄)], min[f-code1(r ₁), f-code0(r ₃)], min[f-code 1(r ₁),f-code2(r ₃)], min [f-code1(r ₁), f-code1(r ₂), f-code1(r ₃)],min[f-code 1(r ₁), f-code2(r ₂), f-code 1(r ₃)]}.

The beneficial effects of the present invention include: the presentinvention is simple and easy to implement, and is especially suitablefor the application of online transformer state monitoring systems;based on a concept of fuzzy logic, diagnosis of composite defects of thetransformer in a complex state and evaluation of the degree of severitycan be achieved, and the problem of sudden change caused by criterionboundary absolutization can be effectively avoided; and a variety ofcharacteristic information, such as alert value and ratio of the gasdissolved in the oil, are merged and analyzed, and thereby thereliability of diagnosis is improved effectively.

DESCRIPTION OF THE DRAWINGS

FIG. 1 is a diagnostic flow chart according to the present invention;

FIG. 2 shows the fuzzy boundary when the ratio code of the ratio r₃ is2.

EMBODIMENTS

Hereunder the present invention will be further described in an example,with reference to FIGS. 1-2.

The implementation steps of this example are as follows:

(I) Acquiring monitoring data of volumetric concentrations of five typesof monitored characteristic gases, i.e., hydrogen, methane, ethane,ethylene and acetylene; calculating the sum of the volumetricconcentrations of methane, ethane, ethylene and acetylene (i.e.,volumetric concentration of total hydrocarbons) from the monitoringdata; judging whether the monitoring data of the five types ofcharacteristic gases or the volumetric concentration of totalhydrocarbons exceeds an alert value; the alert value is selected as perthe Chinese Standard GB/T7252-2001 “Guide to the Analysis and Diagnosisof Gases Dissolved in Transformer Oil”; if the monitoring data or thevolumetric concentration of total hydrocarbons exceeds the alert value,further diagnosis is required; in that case, going to step (II);otherwise judging that the transformer has no defect or fault, if themonitoring data and the volumetric concentration of total hydrocarbonsare normal;

(II) Determining ratio codes:

First, the ratios are set as follows:

${r_{1} = \frac{c_{1}\left( {C_{2}H_{2}} \right)}{c_{2}\left( {C_{2}H_{4}} \right)}},{r_{2} = \frac{c_{3}\left( {CH}_{4} \right)}{c_{4}\left( H_{2} \right)}},{r_{3} = \frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{5}\left( {C_{2}H_{6}} \right)}},{r_{4} = {{\frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{5}\left( {C_{2}H_{6}} \right)}\frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{1}\left( {C_{2}H_{2}} \right)}} = {{r_{3}\frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{1}\left( {C_{2}H_{2}} \right)}}i}}}$

Where, c₁(C₂H₂), c₂(C₂H₄), c₃(CH₄), c₄(H₂), and c₅(C₂H₆) respectivelyrepresent the volumetric concentration of five types of characteristicgases (acetylene, ethylene, methane, hydrogen and ethane), in unit ofμL/L;

Then, the ratio codes are determined according to the rules shown inTable 1:

TABLE 1 Rules for Determining Ratio Codes Ratio Range Ratio Code RatioRange Ratio Code r₁, r₂ or r₃ r₁ r₂ r₃ r₄ r₄ <0.1 0 1 0 ≦1.5 0 ≧0.1~<1 10 0   ≧1~<3 1 2 1 >1.5 1 ≧3 2 2 2

wherein, the ratio codes of r₁, r₂ and r₃ are obtained according to theratio code rules specified in the Chinese Standard GB/T 7252-2001 “Guideto the Analysis and Diagnosis of Gases Dissolved in Transformer Oil”; aratio r₄ and a r₄ ratio code rule are added on the basis of the ratiocode rules specified in the Chinese Standard GB/T 7252-2001 “Guide tothe Analysis and Diagnosis of Gases Dissolved in Transformer Oil”;

(III) Correcting the three-ratio method for judging the types oftransformer defects or faults specified in the Chinese Standard GB/T7252-2001 “Guide to the Analysis and Diagnosis of Gases Dissolved inTransformer Oil” on the basis of analysis of 728 typical real faultcases of the State Grid Corporation of China; thus, rules for judgingthe types of transformer defects or faults based on ratio codes areobtained, as shown in Table 2.

A fourth ratio r₄ is added on the basis of three ratio codes; for thetype of defect or fault with ratio code 101 diagnosed with thethree-ratio method, if r₄≦1.5, the transformer is judged as having aspark discharge defect or fault; if r₄>1.5, the transformer is judged ashaving an arc discharge defect or fault;

Based on the codes for transformer faults and defects specified in theChinese Standard GB/T 7252-2001 “Guide to the Analysis and Diagnosis ofGases Dissolved in Transformer Oil”, a ratio code 011 corresponding topartial discharge fault or defect is added.

TABLE 2 Method for Judging the Types of Transformer Defects or Faultsaccording to Ratio Codes Combinations of Ratio Codes r₁ r₂ r₃ r₄ Type ofDefect or Fault 0 1 0, 1, 2 0, 1 Local discharge 0 1 0, 1Low-temperature overheat lower 2 0 0, 1 than 300□ 2 1 0, 1 300-700° C.moderate-temperature overheat 0, 2 2 0.1 High-temperature overheathigher than 700□ 2 0, 1, 2 0, 1, 2 0, 1 Spark discharge 1 0 1 0 1 Arcdischarge 0, 1, 2 0, 2 0, 1 1, 2 1

(IV) Blurring the boundaries of the codes in Table 1 with a semi-Cauchyrising/falling function, and representing the rising edges and fallingedges of the boundaries with the semi-Cauchy rising/falling function, inorder to change the either-or absolutized boundary judgment; then,obtaining the probabilities of the cases that the ratio codes of theratios r₁, r₂ and r₃ are 0, 1 and 2 respectively (represented byf-code0(r_(i)), f-code1(r_(i)), f-code2(r_(i)) respectively) and theprobabilities of the cases that the ratio code of r4 is 0 or 1respectively, with the semi-Cauchy rising/falling function. For example,the probability of the case that the ratio code of the ratio r₃ is 2 isrepresented by f-code2(r₃), and the fuzzy boundary is represented by thesemi-Cauchy rising edge function, as shown in FIG. 2.

The boundary ranges of the ratios r₁, r₂, r₃ and r₄ are blurred with asemi-Cauchy rising/falling function, and the rising edges and fallingedges of the boundaries are represented with the semi-Cauchyrising/falling function as follows:

${\mu_{d}(r)} = \left\{ {{\begin{matrix}1 & {,{r \leq A}} \\\frac{1}{1 + \left( \frac{A - r}{a} \right)^{2}} & {,{others}}\end{matrix}{\mu_{a}(r)}} = \left\{ \begin{matrix}1 & {,{r \geq A}} \\\frac{1}{1 + \left( \frac{A - r}{a} \right)^{2}} & {,{others}}\end{matrix} \right.} \right.$

wherein, μ_(d)(r) is a falling edge function; μ_(a)(r) is a rising edgefunction; A is a boundary parameter; a is a distribution parameter; thevalues of A and a are optimal values obtained through verification ofthe data of 728 typical real fault cases of the State Grid Coporation ofChina, as shown in Table 3.

TABLE 3 Boundary Parameter A and Distribution Parameter a A₁(r₁) A₂(r₁)A₁(r₂) A₂(r₂) A₁(r₃) A₂(r₃) A(r₄) 0.08 3.1 0.06 0.6 0.8 3.6 1.43  a₁(r₁) a₂(r₁)  a₁(r₂)  a₂(r₂)  a₁(r₃) a₂ (r₃)  a(r₄) 0.01 0.1 0.02 0.2 0.1 0.30.1 

In the Table 3:

The rising edge boundary parameter of r₁, A₁(r₁), is 0.08, and thecorresponding distribution parameter a₁(r₁) is 0.01;

The falling edge boundary parameter of r₁, A₂(r₁), is 3.1, and thecorresponding distribution parameter a₂(r₁) is 0.1;

The rising edge boundary parameter of r₂, A₁(r₂), is 0.06, and thecorresponding distribution parameter a₁(r₂) is 0.02;

The falling edge boundary parameter of r₂, A₂(r₂), is 0.6, and thecorresponding distribution parameter a₂(r₂) is 0.2;

The rising edge boundary parameter of r₃, A₁(r₃), is 0.8, and thecorresponding distribution parameter a₁(r₃) is 0.1;

The falling edge boundary parameter of r₃, A₂(r₃), is 3.6, and thecorresponding distribution parameter a₂(r₃) is 0.3;

The boundary parameter of r₄, A(r₄), is 1.43, and the correspondingdistribution parameter a(r₄) is 0.1;

(V) Obtaining the probabilities of the cases that the ratio codes of theratios r₁, r₂ and r₃ are 0, 1 and 2 respectively and the probabilitiesof the cases that the ratio code of r₄ is 0 or 1 respectively, with thesemi-Cauchy rising/falling function; the expressions are as follows:

Probability f-code0(r₁) of the case that the ratio code of r₁ is 0:

$\begin{matrix}{{f\text{-}{code}\; 0\left( r_{1} \right)} = \left\{ \begin{matrix}1 & \left( {r_{1} \leq 0.08} \right) \\\frac{1}{1 + \left( \frac{0.08 - r_{1}}{0.01} \right)^{2}} & \left( {r_{1} > 0.08} \right)\end{matrix} \right.} & \left( {{expression}\mspace{14mu} 1} \right)\end{matrix}$

Probability f-code1(r₁) of the case that the ratio code of r₁ is 1:

$\begin{matrix}{{f\text{-}{code}\; 1\left( r_{1} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{0.08 - r_{1}}{0.01} \right)^{2}} & \left( {r_{1} \leq 0.08} \right) \\1 & \left( {0.08 \leq r_{1} \leq 3.1} \right) \\\frac{1}{1 + \left( \frac{3.1 - r_{1}}{0.1} \right)^{2}} & \left( {r_{1} > 3.1} \right)\end{matrix} \right.} & \left( {{expression}\mspace{14mu} 2} \right)\end{matrix}$

Probability f-code2(r₁) of the case that the ratio code of r₁ is 2:

$\begin{matrix}{{f\text{-}{code}\; 2\left( r_{1} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{3.1 - r_{1}}{0.1} \right)^{2}} & \left( {r_{1} < 3.1} \right) \\1 & \left( {r_{1} \geq 3.1} \right)\end{matrix} \right.} & \left( {{expression}\mspace{14mu} 3} \right)\end{matrix}$

Probability f-code0(r₂) of the case that the ratio code of r₂ is 0:

$\begin{matrix}{{f\text{-}{code}\; 0\left( {r\; 2} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{0.06 - r_{2}}{0.02} \right)^{2}} & \left( {r_{2} < 0.06} \right) \\1 & \left( {0.06 \leq r_{2} \leq 0.6} \right) \\\frac{1}{1 + \left( \frac{0.6 - r_{2}}{0.2} \right)^{2}} & \left( {r_{2} > 0.6} \right)\end{matrix} \right.} & \left( {{expression}\mspace{14mu} 4} \right)\end{matrix}$

Probability f-code1(r₂) of the case that the ratio code of r₂ is 1:

$\begin{matrix}{{f\text{-}{code}\; 1\left( {r\; 2} \right)} = \left\{ \begin{matrix}1 & \left( {r_{2} \leq 0.06} \right) \\\frac{1}{1 + \left( \frac{0.06 - r_{2}}{0.02} \right)^{2}} & \left( {r_{2} > 0.06} \right)\end{matrix} \right.} & \left( {{expression}\mspace{14mu} 5} \right)\end{matrix}$

Probability f:code2(r₂) of the case that the ratio code of r₂ is 2:

$\begin{matrix}{{f\text{-}{code}\; 2\left( r_{2} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{0.6 - r_{2}}{0.2} \right)^{2}} & \left( {r_{2} < 0.6} \right) \\1 & \left( {r_{2} \geq 0.6} \right)\end{matrix} \right.} & \left( {{expression}\mspace{14mu} 6} \right)\end{matrix}$

Probability f-code0(r₃) of the case that the ratio code of r₃ is 0:

$\begin{matrix}{{f\text{-}{code}\; 0\left( r_{3} \right)} = \left\{ \begin{matrix}1 & \left( {r_{3} \leq 0.8} \right) \\\frac{1}{1 + \left( \frac{0.8 - r_{3}}{0.1} \right)^{2}} & \left( {r_{3} > 0.8} \right)\end{matrix} \right.} & \left( {{expression}\mspace{14mu} 7} \right)\end{matrix}$

Probability f-code1 (r₃) of the case that the ratio code of r₃ is 1:

$\begin{matrix}{{f\text{-}{code}\; 1\left( r_{3} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{0.6 - r_{3}}{0.1} \right)^{2}} & \left( {r_{3} < 0.8} \right) \\1 & \left( {0.8 \leq r_{3} \leq 3.6} \right) \\\frac{1}{1 + \left( \frac{3.6 - r_{3}}{0.3} \right)^{2}} & \left( {r_{3} > 3.6} \right)\end{matrix} \right.} & \left( {{expression}\mspace{14mu} 8} \right)\end{matrix}$

Probability f-code2(r₃) of the case that the ratio code of r₃ is 2:

$\begin{matrix}{{f\text{-}{code}\; 2\left( r_{3} \right)} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{3.6 - r_{3}}{0.3} \right)^{2}} & \left( {r_{3} > 3.6} \right) \\1 & \left( {0.8 \leq r_{3} \leq 3.6} \right)\end{matrix} \right.} & \left( {{expression}\mspace{14mu} 9} \right)\end{matrix}$

Probability f-code0(r₄) of the case that the ratio code of r₄ is 0:

$\begin{matrix}{{f\text{-}{code}\; 0\left( r_{4} \right)} = \left\{ \begin{matrix}1 & \left( {r_{4} \leq 1.43} \right) \\\frac{1}{1 + \left( \frac{1.43 - r_{4}}{0.1} \right)^{2}} & \left( {r_{4} > 1.43} \right)\end{matrix} \right.} & \left( {{expression}\mspace{14mu} 10} \right)\end{matrix}$

Probability f-code1(r₄) of the case that the ratio code of r₄ is 1:

$\begin{matrix}{{f - {{code}\; 1\left( r_{4} \right)}} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{1.43 - r_{4}}{0.1} \right)^{2}} & \left( {r_{4} < 1.43} \right) \\1 & \left( {r_{4} \geq 1.43} \right)\end{matrix} \right.} & \left( {{expression}\mspace{14mu} 11} \right)\end{matrix}$

(VI) Replacing the 0 logic and 1 logic in the ratio code judgement rulewith minimum value logic and maximum value logic respectively, carryingout defect and fault diagnosis according to the correspondence betweenthe ratio codes and the types of transformer defects or faults, andrepresenting the result of diagnosis in a fuzzy multi-value form; theresult is represented in the form of probability, the result ofdiagnosis is the probability of occurrence of defect, i.e., severity;the sum of the probabilities of all kinds of faults is 1; theprobabilities of the ratio codes are represented by maximum value logicand minimum value logic, and the probabilities of the faults are:

f(partial discharge)=min[f-code0(r ₁), f-code1(r ₂)];   (expression 12)

f(low-temperature overheat)=max{min[f-code0(r ₁), f-code0(r ₂),f-code1(r ₃)], min[f-code0(r ₁), f-code2(r ₂), f-code0(r3)]};  (expression 13)

f(moderate-temperature overheat)=min[f-code0(r ₁), f-code2(r ₂),f-code1(r3)];

f(high-temperature overheat)=max{min[f-code0(r ₁), f-code0(r ₂),f-code2(r ₃)], min[f-code0(r ₁), f-code2(r2), f-code2(r ₃)]};  (expression 14)

f(spark discharge)=max{f-code2(r ₁), min[f-code 1(r ₁), f-code0(r ₂),f-code1(r ₃), f-code0(r ₄)]};   (expression 15)

f(arc discharge)=max{min[f-code 1(r ₁), f-code0(r ₂), f-code1(r ₃),f-code1(r ₄)], min[f-code1(r ₁), f-code0(r ₃)], min[f-code 1(r ₁),f-code2(r ₃)], min [f-code1(r ₁), f-code1(r ₂), f-code1(r ₃)],min[f-code 1(r ₁), f-code2(r ₂), f-code 1(r ₃)]}.   (expression 16)

EXAMPLE 1

The oil chromatogram test data of a transformer (volumetricconcentrations of five characteristic gases and total hydrocarbons, inunit of μL/L) is listed in Table 4.

TABLE 4 Oil Chromatogram Test Data of a Transformer c_(z)(total TestDate c₄(H₂) c₃(CH₄) c₅(C₂H₆) c₂(C₂H₄) c₁(C₂H₂) hydrocarbons) Apr. 26,2012 31.33 10.52 1.98 4.01 6.09 22.60

As can be seen from Table 4, the volumetric concentration of acetyleneexceeds the alert value, thus the transformer is abnormal.

-   1. Calculating the four ratios respectively:

${r_{1} = {\frac{c_{1}\left( {C_{2}H_{2}} \right)}{c_{2}\left( {C_{2}H_{4}} \right)} = 1.52}},{r_{2} = {\frac{c_{3}\left( {CH}_{4} \right)}{c_{4}\left( H_{2} \right)} = 2.03}},{r_{3} = {\frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{5}\left( {C_{2}H_{6}} \right)} = 2.03}},{{r_{4} = {{\frac{c_{1}\left( {C_{2}H_{4}} \right)}{c_{5}\left( {C_{2}H_{6}} \right)} \times \frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{1}\left( {C_{2}H_{2}} \right)}} = 1.33}};}$

-   2. Calculating with the expressions 1 to 11, to obtain the    probabilities of the ratio codes of the four ratios:

f-code0(r₁)=0; f-code1(r₁)=1; f-code2(0=0.004;

f-code0(r₂)=1; f-code1(r₂)=0.0051; f-code2(r₂)=0.37;

f-code0(r₃)=0.00657; f-code l(r₃)=1; f-code2(r₃)=0.035;

f-code0(r₄)=-1; f-code2(r₄)=0.5;

-   3. Calculating with the expression 12 to 16, to obtain the    probabilities of the faults:

f(partial discharge)=0%;

f(low-temperature overheat)=0%;

f(moderate-temperature overheat)=0%;

f(high-temperature overheat)=0%;

f(spark discharge)=66.7%;

f(arc discharge)=33.3%;

-   4. Diagnosing the transformer faults

It is judged from the above probabilities of faults, the transformer hasspark discharge fault and arc discharge fault.

While the above embodiments are only preferred embodiments of thepresent invention, the feasible embodiments of the present invention arenot exhausted. Those having ordinary skills in the art may make obviousmodifications without departing from the principle and spirit of thepresent invention; however, all of such modifications shall be deemed asfalling into the scope of protection of the present invention as definedby the claims.

1. A fuzzy diagnosis method for transformer internal composite defectbased on gas dissolved in oil, comprising the following steps: (I)acquiring monitoring data of volumetric concentrations of five types ofmonitored characteristic gases, i.e., hydrogen, methane, ethane,ethylene and acetylene; calculating the sum of the volumetricconcentrations of methane, ethane, ethylene and acetylene (i.e.,volumetric concentration of total hydrocarbons) from the monitoringdata; judging whether the monitoring data of the five types ofcharacteristic gases or the volumetric concentration of totalhydrocarbons exceeds an alert value, which is selected as per theChinese Standard GB/T7252-2001; if the monitoring data or the volumetricconcentration of total hydrocarbons exceeds the alert value, furtherdiagnosis is required; in that case, going to step (II); otherwisejudging that the transformer has no defect or fault, if the monitoringdata and the volumetric concentration of total hydrocarbons are normal;(II) determining ratio codes: first, the ratios are set as follows:${r_{1} = \frac{c_{1}\left( {C_{2}H_{2}} \right)}{c_{2}\left( {C_{2}H_{4}} \right)}},{r_{2} = \frac{c_{3}\left( {CH}_{4} \right)}{c_{4}\left( H_{2} \right)}},{r_{3} = \frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{5}\left( {C_{2}H_{6}} \right)}},{r_{4} = {{\frac{c_{1}\left( {C_{2}H_{2}} \right)}{c_{5}\left( {C_{2}H_{6}} \right)} \times \frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{1}\left( {C_{2}H_{2}} \right)}} = {r_{3} \times \frac{c_{2}\left( {C_{2}H_{4}} \right)}{c_{1}\left( {C_{2}H_{2}} \right)}}}},$wherein, c₁(C₂H₂), c₂(C₂H₄), c₃(CH₄), c₄(H₂) and c₅(C₂H₆) respectivelyrepresent the volumetric concentration of five types of characteristicgases (acetylene, ethylene, methane, hydrogen and ethane), in unit ofμL/L; then, the ratio codes are determined according to the followingrules: if r₁<0.1, the ratio code of r₁ is 0; if 0.15_r₁<1, the ratiocode of r₁ is 1; if 5.1r₁<3, the ratio code of r₁ is 1; if r₁ the ratiocode of r₁ is 2; if r₂<0.1, the ratio code of r₂ is 1; if 0.1≦r₂<1, theratio code of r₂ is 0; if 15-r₂<3, the ratio code of r₂ is 2; if theratio code of r₂ is 2; if r₃<0.1, the ratio code of r₃ is 0; if0.15r₃<1, the ratio code of r₃ is 0; if 1.5r₃<3, the ratio code of r₃ is1; if r₃≧3, the ratio code of r₃ is 2; if r₄≦1.5, the ratio code of r₄is 0; if r₄>1.5, the ratio code of r₄ is 1; (III) Correcting the methodfor determining the types of transformer defects or faults on the basisof three ratios as specified in the Chinese Standard GB/T7252-2001:based on the types of transformer defects or faults corresponding to thethree ratio codes specified in the Chinese Standard GB/T 7252-2001, aratio code 011 corresponding to the type of partial discharge defect orfault is added; a fourth ratio r₄ is added on the basis of the threeratio codes; for the type of defect or fault with ratio code 101diagnosed with the three-ratio method, if the transformer is judged ashaving a spark discharge defect or fault; if r₄>1.5, the transformer isjudged as having an arc discharge defect or fault; thus, obtaining amethod for judging the types of transformer defects or faults accordingto ratio codes as follows: if the ratio code of r₁ is 0, the ratio codeof r₂ is 1, the ratio code of r₃ is 0, 1 or 2, and the ratio code of r₄is 0 or 1, the type of transformer defect or fault is partial discharge;if the ratio code of r₁ is 0, the ratio code of r₂ is 0, the ratio codeof r₃ is 1, and the ratio code of r₄ is 0 or 1, the type of transformerdefect or fault is low-temperature overheat lower than 300□; if theratio code of r₁ is 0, the ratio code of r₂ is 2, the ratio code of r₃is 0, and the ratio code of r₄ is 0 or 1, the type of transformer defector fault is low-temperature overheat lower than 300□; if the ratio codeof r₁ is 0, the ratio code of r₂ is 2, the ratio code of r₃ is 1, andthe ratio code of r₄ is 0 or 1, the type of transformer defect or faultis 300-700° C. moderate-temperature overheat; if the ratio code of r₁ is0, the ratio code of r₂ is 0 or 2, the ratio code of r₃ is 2, and theratio code of r₄ is 0 or 1, the type of transformer defect or fault ishigh-temperature overheat higher than 700□; if the ratio code of r₁ is2, the ratio code of r₂ is 0, 1 or 2, the ratio code of r₃ is 0, 1 or 2,and the ratio code of r₄ is 0 or 1, the type of transformer defect orfault is spark discharge; if the ratio code of r₁ is 1, the ratio codeof r₂ is 0, the ratio code of r₃ is 1, and the ratio code of r₄ is 0,the type of transformer defect or fault is spark discharge; if the ratiocode of r₁ is 1, the ratio code of r₂ is 0, the ratio code of r₃ is 1,and the ratio code of r₄ is 1, the type of transformer defect or faultis arc discharge; if the ratio code of r₁ is 1, the ratio code of r₂ is0, 1 or 2, the ratio code of r₃ is 0 or 2, and the ratio code of r₄ is 0or 1, the type of transformer defect or fault is arc discharge; if theratio code of r₁ is 1, the ratio code of r₂ is 1 or 2, the ratio code ofr₃ is 1, and the ratio code of r₄ is 0 or 1, the type of transformerdefect or fault is arc discharge; (IV)blurring the boundary ranges ofthe ratios r₁, r₂, r₃ and r₄ with a semi-Cauchy rising/falling function,and representing the rising edges and falling edges of the boundarieswith the semi-Cauchy rising/falling function as follows:${\mu_{a}(r)} = \left\{ {{\begin{matrix}{1,} & {r \leq A} \\{\frac{1}{1 + \left( \frac{A - r}{a} \right)^{2}},} & {others}\end{matrix}{\mu_{a}(r)}} = \left\{ \begin{matrix}{1,} & {r \geq A} \\{\frac{1}{1 + \left( \frac{A - r}{a} \right)^{2}},} & {others}\end{matrix} \right.} \right.$ wherein, μ_(d)(r) is a falling edgefunction; μ_(a)(r) is a rising edge function; A is a boundary parameter;a is a distribution parameter; the values of A and a are as follows: therising edge boundary parameter of r₁ is 0.08, and the correspondingdistribution parameter is 0.01, the falling edge boundary parameter ofr₁ is 3.1, and the corresponding distribution parameter is 0.1; therising edge boundary parameter of r₂ is 0.06, and the correspondingdistribution parameter is 0.02; the falling edge boundary parameter ofr₂ is 0.6, and the corresponding distribution parameter is 0.2; therising edge boundary parameter of r₃ is 0.8, and the correspondingdistribution parameter is 0.1; the falling edge boundary parameter of r₃is 3.6, and the corresponding distribution parameter is 0.3; theboundary parameter of r₄ is 1.43, and the corresponding distributionparameter is 0.1; (V) obtaining the probabilities of the cases that theratio codes of the ratios r₁, r₂ and r₃ are 0, 1 and 2 respectively andthe probabilities of the cases that the ratio code of r₄ is 0 or 1respectively, with the semi-Cauchy rising/falling function; theexpressions are as follows: Probability f-code0(r₁) of the case that theratio code of r₁ is 0:${f - {{code0}\; \left( r_{1} \right)}} = \left\{ \begin{matrix}1 & \left( {r_{1} \leq 0.08} \right) \\\frac{1}{1 + \left( \frac{0.08 - r_{1}}{0.01} \right)^{2}} & \left( {r_{1} > 0.08} \right)\end{matrix} \right.$ Probability f-code1(r₁) of the case that the ratiocode of r₁ is 1:${f - {{code}\; 1\left( r_{1} \right)}} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{0.08 - r_{1}}{0.01} \right)^{2}} & \left( {r_{1} \leq 0.08} \right) \\1 & \left( {0.08 \leq r_{1} \leq 3.1} \right) \\\frac{1}{1 + \left( \frac{3.1 - r_{1}}{0.1} \right)^{2}} & \left( {r_{1} > 3.1} \right)\end{matrix} \right.$ probability f-code2(r₁) of the case that the ratiocode of r₁ is 2:${f - {{code}\; 2\; \left( r_{1} \right)}} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{3.1 - r_{1}}{0.1} \right)^{2}} & \left( {r_{1} < 3.1} \right) \\1 & \left( {r_{1} \geq 3.1} \right)\end{matrix} \right.$ probability f-code0(r₂) of the case that the ratiocode of r₂ is 0:${f - {{code}\; 0\left( r_{2} \right)}} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{0.06 - r_{2}}{0.02} \right)^{2}} & \left( {r_{2} < 0.06} \right) \\1 & \left( {0.06 \leq r_{2} \leq 0.6} \right) \\\frac{1}{1 + \left( \frac{0.06 - r_{2}}{0.02} \right)^{2}} & \left( {r_{2} > 0.6} \right)\end{matrix} \right.$ probability f-code1(r₂) of the case that the ratiocode of r₂ is 1:${f - {{code}\; 1\; \left( {r\; 2} \right)}} = \left\{ \begin{matrix}1 & \left( {r_{2} \leq 0.06} \right) \\{1 + \left( \frac{0.06 - r_{2}}{0.02} \right)^{2}} & \left( {r_{2} > 0.06} \right)\end{matrix} \right.$ probability f-code2(r₂) of the case that the ratiocode of r₂ is 2:${f - {{code}\; 2\; \left( r_{2} \right)}} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{0.6 - r_{2}}{0.2} \right)^{2}} & \left( {r_{2} < 0.6} \right) \\1 & \left( {r_{2} \geq 0.6} \right)\end{matrix} \right.$ probability f-code0(r₃) of the case that the ratiocode of r₃ is 0:${f - {{code}\; 0\; \left( r_{3} \right)}} = \left\{ \begin{matrix}1 & \left( {r_{3} \leq 0.8} \right) \\\frac{1}{1 + \left( \frac{0.8 - r_{3}}{0.1} \right)^{2}} & \left( {r_{3} > 0.8} \right)\end{matrix} \right.$ probability f-code1(r₃) of the case that the ratiocode of r₃ is 1:${f - {{code}\; 1\left( r_{3} \right)}} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{0.6 - r_{3}}{0.1} \right)^{2}} & \left( {r_{3} < 0.8} \right) \\1 & \left( {0.8 \leq r_{3} \leq 3.6} \right) \\\frac{1}{1 + \left( \frac{3.6 - r_{3}}{0.3} \right)^{2}} & \left( {r_{3} > 3.6} \right)\end{matrix} \right.$ probability f-code2(r₃) of the case that the ratiocode of r₃ is 2:${f - {{code}\; 2\; \left( r_{3} \right)}} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{3.6 - r_{3}}{0.3} \right)^{2}} & \left( {r_{3} > 3.6} \right) \\1 & \left( {0.8 \leq r_{3} \leq 3.6} \right)\end{matrix} \right.$ probability f-code0(r₄) of the case that the ratiocode of r₄ is 0:${f - {{code}\; 0\; \left( r_{4} \right)}} = \left\{ \begin{matrix}1 & \left( {r_{4} \leq 1.43} \right) \\\frac{1}{1 + \left( \frac{1.43 - r_{4}}{0.1} \right)^{2}} & \left( {r_{4} > 1.43} \right)\end{matrix} \right.$ Probability f-code1(r₄) of the case that the ratiocode of r₄ is 1:${f - {{code}\; 1\; \left( r_{4} \right)}} = \left\{ \begin{matrix}\frac{1}{1 + \left( \frac{1.43 - r_{4}}{0.1} \right)^{2}} & \left( {r_{4} < 1.43} \right) \\1 & \left( {r_{4} \geq 1.43} \right)\end{matrix} \right.$ (VI)representing the probabilities of ratio codeswith maximum value logic and minimum value logic, and thereby obtaininga fuzzy multi-value form of the diagnostic result of the types oftransformer defects or faults; the probabilities of the types oftransformer defects or faults are as follows:f(partial discharge)=min[f-code0(r ₁), f-code1(r₂)];f(low-temperature overheat)=max{min[f-code0(r₁), f-code0(r₂),f-code1(r₃)], min[f-code0(r₁),f-code2(r ₂), f-code0(r3)]};f(moderate-temperature overheat)=min[f-code0(r ₁), f-code2(r ₂),f-code1(r3)];f(high-temperature overheat)=max{min[f-code0(r ₁), f-code0(r ₂),f-code2(r ₃)], min[f-code0(r ₁), f-code2(r2), f-code2(r ₃)]};f(spark discharge)=max{f-code2(r ₁), min[f-code 1(r ₁), f-code0(r ₂),f-code1(r ₃), f-code0(r ₄)]};f(arc discharge)=max{min[f-code 1(r ₁), f-code0(r ₂), f-code1(r ₃),f-code1(r ₄)], min[f-code1(r ₁), f-code0(r ₃)], min[f-code 1(r ₁),f-code2(r ₃)], min [f-code1(r ₁), f-code1(r ₂), f-code1(r ₃)],min[f-code 1(r ₁), f-code2(r ₂), f-code 1(r ₃)]}.